Question

我正在编写一些代码来查找字符串中最长的回文。我想从索引0开始，然后将子字符串的增加长度推送到数组：

ex:
string = "ababa"
[["a", "b"], ["a", "b", "a"], ["a", "b", "a", "b"], ["a", "b", "a", "b", "a"]]

然后它应该从索引1开始并执行相同的操作：

ex:
string = "ababa"
[["b","a"],["b","a","b"],["b","a","b","a"]

这应该一直持续到索引的长度为-1。但是，以下代码在经历了从第一个索引开始的所有迭代后停止并且只返回：

[["a", "b"], ["a", "b", "a"], ["a", "b", "a", "b"], ["a", "b", "a", "b", "a"]]

我逻辑中的缺陷是什么？代码

def longest_palindrome(s)
  array = s.chars
  start = 0
  place = 1
  output = []
  while start < s.length - 1
    while place < s.length
      output << array[start..place]
      place += 1
    end
  start += 1
  end
return output  
end

Answer 1

我相信这就是你所追求的：

def longest_palindrome(s)
  arr = s.chars
  output = []

  (0...s.length).each do |start|
    (start + 1...s.length).each do |place|
      output << arr[start..place]
    end
  end

  output
end

longest_palindrome("ababa")
 => [["a", "b"], ["a", "b", "a"], ["a", "b", "a", "b"], ["a", "b", "a", "b", "a"], ["b", "a"], ["b", "a", "b"], ["b", "a", "b", "a"], ["a", "b"], ["a", "b", "a"], ["b", "a"]]

使用迭代器确实简化了事情。这是一个更简洁的版本：

def longest_palindrome(s)
  output = (0...s.length).flat_map do |start|
    (start + 1...s.length).map do |place|
      s[start..place]
    end
  end
end

longest_palindrome("ababa")
 => ["ab", "aba", "abab", "ababa", "ba", "bab", "baba", "ab", "aba", "ba"]

Answer 2

工作，非优化，即时写入30分钟，谁知道最慢的最长回文查找器：

def longest_palindrome(string, min_size = 2)
  string = string.downcase # we will not be taking 
  return string if string == string.reverse # skip all calculaions of the passed string itself is a palindrome in original order
  letters = string.chars # Convert string to array of bytes
  combinations = [] # Initialize all letter combinations
  (min_size..letters.size).each do |n| # min_size is the shortest length a palindrome is allowed to be, default 2
    combinations.concat(letters.combination(n).to_a) # concat all combinations for n amount of characters in the string
  end
  palindromes = [] # Initialize array for all palindromes
  combinations.each do |combo| # interate every combo
    combo.size.times do # for every letter ordering is done via size of this letter combo
      palindromes << combo.dup if combo == combo.reverse # add to list of palindromes if the combinations is the same backwards
      combo.rotate! # rotate the letters for next order checking
    end
  end
  palindromes.sort {|a, b| a.size <=> b.size }.last.join # sort the palidromes by length, take the biggest one, and return it as a full string
end

p longest_palindrome("racecar") #=> racecar
p longest_palindrome("applesauce") #=> pecep
p longest_palindrome("madam im adam") #=> mada m adam
p longest_palindrome("madamimadam") #=> madamimadam

但它有效。正如你所看到的那样，当混合中添加空格/条纹时，是否会变得奇怪。如果需要，您可以随时在第一行清除它的字符串。

Viva la无算法编码！

红宝石最长的回文。嵌套while循环

2 个答案: